490. The Maze - Explanation

Description

There is a ball in a maze with empty spaces (represented as 0) and walls (represented as 1). The ball can go through the empty spaces by rolling up, down, left or right, but it won't stop rolling until hitting a wall. When the ball stops, it could choose the next direction.

Given the m x n maze, the ball's start position and the destination, where start = [start_row, start_col] and destination = [destination_row, destination_col], return true if the ball can stop at the destination, otherwise return false.

You may assume that the borders of the maze are all walls (see examples).

Example 1:

Input: maze = [[0,0,1,0,0],[0,0,0,0,0],[0,0,0,1,0],[1,1,0,1,1],[0,0,0,0,0]], start = [0,4], destination = [4,4]

Output: true

Explanation: One possible way is : left -> down -> left -> down -> right -> down -> right.

Example 2:

Input: maze = [[0,0,1,0,0],[0,0,0,0,0],[0,0,0,1,0],[1,1,0,1,1],[0,0,0,0,0]], start = [0,4], destination = [3,2]

Output: false

Explanation: There is no way for the ball to stop at the destination. Notice that you can pass through the destination but you cannot stop there.

Example 3:

Input: maze = [[0,0,0,0,0],[1,1,0,0,1],[0,0,0,0,0],[0,1,0,0,1],[0,1,0,0,0]], start = [4,3], destination = [0,1]

Output: false

Constraints:

m == maze.length
n == maze[i].length
1 <= m, n <= 100
maze[i][j] is 0 or 1.
start.length == 2
destination.length == 2
0 <= start_row, destination_row < m
0 <= start_col, destination_col < n
Both the ball and the destination exist in an empty space, and they will not be in the same position initially.
The maze contains at least 2 empty spaces.

Company Tags

1. Depth First Search

class Solution:
    def dfs(self, m, n, maze, curr, destination, visit):
        if visit[curr[0]][curr[1]]:
            return False
        if curr[0] == destination[0] and curr[1] == destination[1]:
            return True

        visit[curr[0]][curr[1]] = True
        dirX = [0, 1, 0, -1]
        dirY = [-1, 0, 1, 0]

        for i in range(4):
            r = curr[0]
            c = curr[1]
            # Move the ball in the chosen direction until it can.
            while r >= 0 and r < m and c >= 0 and c < n and maze[r][c] == 0:
                r += dirX[i]
                c += dirY[i]
            # Revert the last move to get the cell to which the ball rolls.
            if self.dfs(m, n, maze, [r - dirX[i], c - dirY[i]], destination, visit):
                return True
        return False

    def hasPath(self, maze: List[List[int]], start: List[int], destination: List[int]) -> bool:
        m = len(maze)
        n = len(maze[0])
        visit = [[False] * n for _ in range(m)]
        return self.dfs(m, n, maze, start, destination, visit)

Time & Space Complexity

Time complexity: $O(m \cdot n \cdot (m + n))$
Space complexity: $O(m \cdot n)$

Where $m$ and $n$ are the number of rows and columns in maze.

2. Breadth First Search

class Solution:
    def hasPath(self, maze: List[List[int]], start: List[int], destination: List[int]) -> bool:
        m = len(maze)
        n = len(maze[0])
        visit = [[False] * n for _ in range(m)]
        queue = deque()
        
        queue.append(start)
        visit[start[0]][start[1]] = True
        dirX = [0, 1, 0, -1]
        dirY = [-1, 0, 1, 0]

        while queue:
            curr = queue.popleft()
            if curr[0] == destination[0] and curr[1] == destination[1]:
                return True

            for i in range(4):
                r = curr[0]
                c = curr[1]
                # Move the ball in the chosen direction until it can.
                while r >= 0 and r < m and c >= 0 and c < n and maze[r][c] == 0:
                    r += dirX[i]
                    c += dirY[i]
                # Revert the last move to get the cell to which the ball rolls.
                r -= dirX[i]
                c -= dirY[i]
                if not visit[r][c]:
                    queue.append([r, c])
                    visit[r][c] = True
        return False

Time & Space Complexity

Time complexity: $O(m \cdot n \cdot (m + n))$
Space complexity: $O(m \cdot n)$

Where $m$ and $n$ are the number of rows and columns in maze.